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Kamila [148]
3 years ago
8

Point G is the centroid of the right △ABC with hypotenuse AB=18 in. Find CG.

Mathematics
1 answer:
Rama09 [41]3 years ago
5 0
<h3>Answer:</h3>

6 in

<h3>Explanation:</h3>

Let M be the midpoint of AB. ∆CMB and ∆CMA are both isosceles triangles, so CM = (AB)/2 = 9 in.

CG = (2/3)(CM) = (2/3)(9 in)

CG = 6 in

_____

<em>Comment on centroid and median</em>

The centroid of a triangle is located 1/3 the distance from the midpoint of a side to the opposite vertex. This is true for any median in any triangle. The proof can be developed from the fact that every median divides the triangle's area in half. Here, we just take advantage of this fact.

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Find the general term that represents the situation in terms of k.

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a_{n}=a_{1}r^{n-1}

a_{1} = the first term of the series

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a_{1} would represent the height at which the ball is first dropped. Therefore:

a_{1} = k

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r=\frac{3}{4}

Our general term would be:

a_{n}=a_{1}r^{n-1}

a_{n}=k(\frac{3}{4}) ^{n-1}

Second part of question:

If the ball dropped from a height of 235ft, determine the highest height achieved by the ball after six bounces.

k represents the initial height:

k = 235\ ft

n represents the number of times the ball bounces:

n = 6

Plugging this back into our general term of the geometric series:

a_{n}=k(\frac{3}{4}) ^{n-1}

a_{n}=235(\frac{3}{4}) ^{6-1}

a_{n}=235(\frac{3}{4}) ^{5}

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Third part of question:

If the ball dropped from a height of 235ft, find the total distance traveled by the ball when it strikes the ground for the 12th time. ​

This would be easier to solve if we have a general term for the <em>sum </em>of a geometric series, which is:

S_{n}=\frac{a_{1}(1-r^{n})}{1-r}

We already know these variables:

a_{1}= k = 235\ ft

r=\frac{3}{4}

n = 12

Therefore:

S_{n}=\frac{(235)(1-\frac{3}{4} ^{12})}{1-\frac{3}{4} }

S_{n}=\frac{(235)(1-\frac{3}{4} ^{12})}{\frac{1}{4} }

S_{n}=(4)(235)(1-\frac{3}{4} ^{12})

S_{n}=910.22\ ft

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